- #1
CAF123
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Homework Statement
[/B]
Consider the missing information function ##S(\left\{p_i\right\}) = -k \sum_{i=1}^r p_i \ln p_i## where ##\left\{p_i\right\}## represents the probabilities associated with the ##r## mutually exclusive outcomes of some procedure.
i) Sketch the form of S as a function of ##p_1## for ##r=2##
ii) Show that ##S## has its max value when ##p_i = 1/r \forall i##
iii) Show that the value of ##S## is increased when any two of the probablities are changed in such a way as to reduce their difference.
2. Homework Equations
3. The Attempt at a Solution
In i), we take ##S = S(p_1, p_2)## and regard ##p_2## as a constant. Then ##p_1 \in [0,1]## is the domain and the plot of S versus p_1 looks like a skewed hump with max at ##p_1 = 1/e## and ##S=0## when ##p_1 = 0,1##. Did I interpret the question here right? I just don't see why we restrict ##r## to 2.
ii) is fine I think. I get that ##p_j = 1/e## then ##\sum_{i=1}^r p_j = 1 \Rightarrow r (1/e) = 1 = r p_j \Rightarrow p_j = 1/r##. What is the significance of 1/e though? The result wouldn't change as long as it was a constant.
In iii), not really sure where to start. I think the condition is that if ##|p_i + \delta - (p_j + \gamma)| < |p_i - p_j|## then ##dS > 0##, where ##\delta, \gamma## are real numbers. But not sure how to implement this.